Yes. If

g

generates the cyclic group

C_{n}

ρ (g) = e^{2 πi / n} .

It is irreducible because it is one-dimensional, and it is faithful because the displayed scalar has multiplicative order

n

. Thus every

C_{n}

, including the trivial case

n = 1

ii

Solution

Yes. Write

D_{8} = ⟨ r, s : r^{4} = s^{2} = 1, srs = r^{- 1} ⟩ .

ρ (r) = (i 0 0 - i), ρ (s) = (0110) .

These matrices satisfy the defining relations. The two eigenspaces of

ρ (r)

are exchanged by

ρ (s)

, so there is no common one-dimensional invariant subspace; the representation is irreducible. Its eight matrices

ρ (r)^{j}

and

ρ (r)^{j} ρ (s)

are distinct, so it is faithful.

No. The center of a group satisfies

Z (C_{2} \times D_{8}) = C_{2} \times Z (D_{8}) ≅ C_{2} \times C_{2},

because

Z (D_{8}) = {1, r^{2}} ≅ C_{2}

C_{2} \times D_{8}

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